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First note that if \(a \sqsubseteq c\) and \(b \sqsubseteq c\), then, since \(c \in A(a) \cap B(b)\), the special property of \(\sqsubseteq\) gives \(a \sim c \sim b\). Now consider the case when \(a \sqsubseteq b\). If \(b \sqsubseteq c\), then by the above remarks \(a \sim c \sim b\). This implies \(g(a)\) is a joinand of \(g(b)\), and so \(g(a) \le g(b)\). Hence we may assume \(c \sqsubset b\), and thus by induction \(g(a) \le g(c)\). Also \(g(c)\) is a joinand of \(g(b)\) and so \(g(c) \le g(b)\) and hence \(g(a) \le g(b)\).